WHomework set 10_solns_2011

WHomework set 10_solns_2011 - Homework set # 10 Problems:...

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Homework set # 10 Problems: 20.9, 10, 12, 17, 21, 22, 23, 33 Solutions: 20.9 From cos B BA tt    , we find that  3 2 18 10 V 0.18 T cos 0.10 m cos0 B At s 20.10   cos B NBA    , so cos t B NA  or     3 5 2 0 . 1 6 6 V 2 . 7 71 0 s 5.20 10 T 52.0 T 500 0.150 m 4 cos0 cos90 B  20.12 The initial magnetic field inside the solenoid is 73 0 100 4 10 T m A 3.00 A 1.88 10 T 0.200 m Bn I      (a)    2 32 72 cos 1.88 10 T 1.00 10 m cos0 1.88 10 T m B BA   (b) When the current is zero, the flux through the loop is 0 B and the average induced emf has been 8 0 6.28 10 V 3.00 s B t  20.17 B v , where B is the component of the magnetic field perpendicular to the velocity v . Thus,   6 50.0 10 T sin58.0 60.0 m 300 m s 0.763 V
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20.21 (a)
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This note was uploaded on 05/12/2011 for the course PHYS 1112 taught by Professor Seaton during the Spring '08 term at UGA.

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WHomework set 10_solns_2011 - Homework set # 10 Problems:...

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